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that's my circuit, but there is another version somewhere where i put the maths in to calculate the component values. The maximum output current, is 1.22/r, which with the componentS shown, would be 2.4 amps. The current

yes i know that but not really good with schematics i guess im askinh the wrong question what im asking is how do i hook up the pot meter to the LM317 as in does the Iadjust solder to the Vout or Iadjust or and dont the LEDS connect to the Vout or adjust

The circuit diagram you attached shows where the components are attached to one another.

The values of these resistors,including the setting of the variable resistor, determine how the circuit will work, specifically how much current it will allow through the load, the series string of LEDs.

The part confusing to me, is you want to substitute different components, like a 10K ohm pot in place of a 500 ohm pot, and then naively expect the circuit to work the same.

It's not like the circuit depends only on where you put the resistors. The values of the resistors matter too.

Consider again the more simple circuit, with just one small resistor R1, and Iout = (1.25 V)/R1; e.g. (1.25 V)/(2.2. ohm) = 0.568 A. This more simple circuit is not adjustable by turning a pot, but it will work the way you expect it to.

im trying to do this and i do have the 500ohm pot but wanted to use the 10k pot i tried getting a hold of maker but he seems lost and not found my only question is where do the leads of the pot meter go or how would i wire it please if you know where i will appericate it thanks in advance

by the way he does not use the resisters in parallel as you see in the pic of the breadboard thats what confuses me

LEDS used all 0.7mA: 6 x Red 630nm 3 x Blue 460nm 2 x Deep Red 660nm

for the constant current driver 2 x LM317 2 x 1.8 ohm resistors, 2.2ohm can also be used if you wanna be more safe 2 x pot-meters 500 ohm 2 x 1000ohm resistors

For this circuit, a constant current source based on LM317 regulator, the math is easy when there is just one small feedback resistor R1 placed between Vout and Vadj. For that easy case, Iout = (1.25 V)/R1 + Iadj, where Iadj is small, typically 50 microamperes, usually small enough to ignore, so Iout = (1.25 V)/R1

For a circuit with a more complicated feedback network, like the circuit you posted, with three resistors, {R1, R2, R3}, one of which is variable resistor (also called potentiometer), the math becomes more complicated, as I am interpreting it.